Prove from the definition that $\frac{x}{1+x}$ is continuous at $x = 1$.

Question

The definition for continuous functions is pretty straightforward, so I wrote:

$|\frac{x}{x+1}-\frac{1}{2}| = |\frac{x-1}{2x+2}| < \epsilon$ if

$|x-1| < \delta$ and $\epsilon, \delta > 0.$

Unfortunately I am not really strong when it comes to epsilon-delta proofs, so any ideas as to how I may proceed?

Thank you!

Answer 1

Suppose $\varepsilon$ is a positive real number. The goal is to find a positive real number $\delta$ such that for every $x \in (1 - \delta, 1 + \delta)$, we have \begin{equation*} \varepsilon > \left| \frac x {1 + x} - \frac 1 2 \right| = \frac {|2x - (1 + x)|} {2|1 + x|} = \frac {|x - 1|} {2|1 + x|}. \end{equation*} Given $x \in (1 - \delta, 1 + \delta)$, we have $|x - 1| < \delta$. Also, if $\delta$ is chosen to be less than 1 then $x$ is positive, so $|1 + x| = 1 + x$ is greater than or equal to 1, and hence the fraction $|x - 1|/2|1 + x|$ is less than its numerator $|x - 1|$.

Therefore, if we choose $\delta = \min(1, \varepsilon)$, then for every $x \in (1 - \delta, 1 + \delta)$, we have \begin{equation*} \left| \frac x {1 + x} - \frac 1 2 \right| = \frac {|2x - (1 + x)|} {2|1 + x|} < |x - 1| < \delta < \varepsilon. \end{equation*}

(The paragraph immediately above is the proof, the paragraph above that is the explanation of how one would come up with this proof.)

Answer 2

Hint: if $\delta<1$ then $x>0$, so what can you say about the denominator, $|2x+2|$?